Consider the function
f (x) = 2x^3 - 3x^2 - 72x + 7
Find and classify the critical points. Identify the intervals of increase and decrease, and state the intervals of concavity.
f(x) = 2x^3 - 3x^2 - 72x + 7
f'(x) = 6x^2 - 6x - 72 = 6(x^2-x-12)
f"(x) = 12x-6 = 6(2x-1)
f'=0 at x= -3 and 4
f"=0 at x = 1/2
Now it's easy to answer the questions.
f is increasing where f' > 0
f is concave up where f" > 0
no asymptotes, so critical points where f' = 0
To find the critical points of a function, we need to first find the derivative of the function and set it equal to zero.
Given f(x) = 2x^3 - 3x^2 - 72x + 7, let's find its derivative, f'(x).
f'(x) = d/dx (2x^3) - d/dx (3x^2) - d/dx (72x) + d/dx (7)
Taking the derivative term by term:
f'(x) = 6x^2 - 6x - 72
Now, we'll set f'(x) equal to zero and solve for x to find the critical points:
6x^2 - 6x - 72 = 0
Dividing both sides by 6:
x^2 - x - 12 = 0
Factoring the quadratic equation:
(x - 4)(x + 3) = 0
Setting each factor equal to zero:
x - 4 = 0 or x + 3 = 0
Solving for x:
x = 4 or x = -3
So, the critical points are x = 4 and x = -3.
To classify the critical points, we need to analyze the sign of the derivative on either side of each critical point.
Let's evaluate f'(x) for values less than -3, between -3 and 4, and greater than 4.
For x < -3:
Plugging in a value such as x = -5, we get:
f'(-5) = 6*(-5)^2 - 6*(-5) - 72 = 150 + 30 - 72 = 108
Since f'(-5) > 0, the derivative is positive, indicating that f(x) is increasing.
For -3 < x < 4:
Plugging in a value such as x = 0, we get:
f'(0) = 6*0^2 - 6*0 - 72 = -72
Since f'(0) < 0, the derivative is negative, indicating that f(x) is decreasing.
For x > 4:
Plugging in a value such as x = 6, we get:
f'(6) = 6*6^2 - 6*6 - 72 = 180 - 36 - 72 = 72
Since f'(6) > 0, the derivative is positive, indicating that f(x) is increasing.
Therefore, we have:
- f(x) is increasing for x < -3 and x > 4.
- f(x) is decreasing for -3 < x < 4.
To find the intervals of concavity, we need to find the second derivative of the function, f''(x).
f''(x) = d/dx(6x^2 - 6x - 72)
Taking the derivative:
f''(x) = 12x - 6
Now, we need to find the values of x where f''(x) = 0 to determine the points of inflection.
12x - 6 = 0
Dividing by 6:
2x - 1 = 0
Solving for x:
x = 1/2
So, x = 1/2 is the point of inflection.
To classify the concavity intervals, we need to evaluate f''(x) for values less than 1/2 and greater than 1/2.
For x < 1/2:
Plugging in a value such as x = 0, we get:
f''(0) = 12*0 - 6 = -6
Since f''(0) < 0, the second derivative is negative, indicating that f(x) is concave downward.
For x > 1/2:
Plugging in a value such as x = 1, we get:
f''(1) = 12*1 - 6 = 6
Since f''(1) > 0, the second derivative is positive, indicating that f(x) is concave upward.
Therefore, we have:
- f(x) is concave downward for x < 1/2.
- f(x) is concave upward for x > 1/2.
In summary:
- The critical points are x = 4 and x = -3.
- f(x) is increasing for x < -3 and x > 4.
- f(x) is decreasing for -3 < x < 4.
- f(x) is concave downward for x < 1/2.
- f(x) is concave upward for x > 1/2.
To find the critical points of the function, we need to find the values of x where the derivative is equal to zero or undefined. Let's start by finding the derivative of the function f(x).
f(x) = 2x^3 - 3x^2 - 72x + 7
To find the derivative, we can use the power rule and the sum/difference rule.
The power rule states that the derivative of x^n is n*x^(n-1), where n is a constant.
The sum/difference rule states that the derivative of the sum or difference of two functions is equal to the sum or difference of their derivatives.
Using these rules, the derivative of f(x) can be found as follows:
f'(x) = 3*2x^2 - 2*3x - 72
= 6x^2 - 6x - 72
= 6(x^2 - x - 12)
Now we need to find the critical points by setting the derivative equal to zero and solving for x:
6(x^2 - x - 12) = 0
Factoring the quadratic equation, we get:
6(x - 4)(x + 3) = 0
Setting each factor equal to zero and solving for x, we find the critical points:
x - 4 = 0 --> x = 4
x + 3 = 0 --> x = -3
So, the critical points of the function are x = 4 and x = -3.
To classify the critical points, we need to evaluate the second derivative of the function at these points.
To find the second derivative, we differentiate the derivative we found earlier:
f''(x) = 6(2x - 1)
Now we can plug in the values of x we found earlier to determine the concavity:
f''(4) = 6(2(4) - 1) = 6(8 - 1) = 6(7) = 42
f''(-3) = 6(2(-3) - 1) = 6(-6 - 1) = 6(-7) = -42
Since f''(4) > 0, the function is concave up at x = 4.
Since f''(-3) < 0, the function is concave down at x = -3.
Now we can determine the intervals of increase and decrease by considering the signs of the derivative.
The derivative is given by:
f'(x) = 6x^2 - 6x - 72
To find the intervals of increase and decrease, we need to analyze the sign of the derivative.
Setting the derivative equal to zero and solving for x, we get:
6x^2 - 6x - 72 = 0
Dividing by 6, we get:
x^2 - x - 12 = 0
Factoring the quadratic equation, we have:
(x - 4)(x + 3) = 0
Setting each factor equal to zero and solving for x, we find:
x = 4 or x = -3
These are the critical points we found earlier.
Now we can create a sign chart to determine the intervals of increase and decrease:
-3 4
|------------|-------------|
+ | (-) | (+) |
f'(x) | | | | |
- | (+) | (-) |
|------------|-------------|
From the sign chart, we see that the function is increasing on the interval (-∞, -3) and decreasing on the interval (-3, 4). Finally, the function is increasing again on the interval (4, ∞).
Therefore, the critical points of the function are x = 4 and x = -3. The intervals of increase are (-∞, -3) and (4, ∞), and the interval of decrease is (-3, 4). The intervals of concavity are (4, ∞) for concave up and (-∞, -3) for concave down.